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    what happens when sulphuric acid is added to sodium iodide?

    initially, the HI produced is a strong reducing agent and reduces H2SO4 and oxidises itself, to iodine (violet iodine vapour). But what is the overall equation?
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    sorry, i've just looked at the EDEXCEL syllabus, and it says the equation for sulphuric with iodides is not required, but what is the answer?
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    (Original post by med115)
    what happens when sulphuric acid is added to sodium iodide?

    initially, the HI produced is a strong reducing agent and reduces H2SO4 and oxidises itself, to iodine (violet iodine vapour). But what is the overall equation?
    This is a nasty one! There are three sets of reactions that go on simultaneously:

    2NaI(s) ---> I2(s) + 2Na+ + 2e-

    H2SO4(l) + 2H+ + 2e- ---> SO2(g) + 2H2O(l)


    6NaI(s) ---> 3I2(s) + 6Na+ + 6e-

    H2SO4(l) + 6H+ + 6e- ---> S(s) + 4H2O(l)


    8NaI(s) ---> 4I2(s) + 8Na+ + 8e-

    H2SO4(l) + 8H+ + 8e- ---> H2S(g) + 4H2O(l)


    I remember doing this experiment. The H2S smells rotten.
 
 
 

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